Matrices Test

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Matrices Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=\begin{bmatrix} 2 & x-3 & x-2 \\ 3 & -2 & -1 \\ 4 & -1 & -5 \end{bmatrix}$$ is a symmetric matrices then $$x=$$

A
$$0$$

B
$$3$$

C
$$6$$

D
$$8$$

Solution

If $$A$$ is a symmetric matrix then $$a_{ij}=a_{ji}$$
i.e. $$a_{12}=a_{21}$$ and $$a_{13}=a_{31}$$
$$\therefore x-3=3$$ and $$x-2=4$$
adding both we get$$ 2x-5=7$$
$$\Rightarrow 2x=12 \Rightarrow x=6$$
Question 2
1 / -0

Given $$A$$ is a matrix of order $$3\times 2$$. If order of $$AB$$ is $$3\times 3$$, then order of $$B$$ will be

A
$$1\times 3$$

B
$$2\times 3$$

C
$$3\times 3$$

D
$$2\times 2$$

Solution

Given $$A$$ is a matrix of order $$3\times 2$$.
Now $$AB$$ is of order $$3\times 3$$.
The product $$AB$$ is possible if the number of columns of $$A(=2)$$ is equal to the number of rows of $$B$$.
Since order of $$AB$$ is $$3\times 3$$.
Also the number of columns of $$B$$ is same as the number of columns of $$AB$$.
Now $$A$$ is of order $$2\times 3$$.
Question 3
1 / -0

Given $$A= \begin{bmatrix} 3&4 \\ 4&-3 \end{bmatrix}$$ and $$B = \begin{bmatrix} 24 \\ 7\end{bmatrix},$$ find the matrix $$X$$ such that $$AX=B$$.

A
$$\begin{bmatrix} -4 \\ -3\end{bmatrix}$$

B
$$\begin{bmatrix} 4 \\ 3\end{bmatrix}$$

C
$$\begin{bmatrix} -4 \\ 3\end{bmatrix}$$

D
$$\begin{bmatrix} 4 \\ -3\end{bmatrix}$$

Solution

Given, $$AX=B$$

To perform matrix multiplication, $$AX$$ the number of columns in A must equal the number of rows in $$X$$

Hence, $$X$$ should be matrix of order $$2\times1$$

Say, $$X= \begin{bmatrix}x \\y\end{bmatrix}$$

Consider, $$AX=B$$

$$\begin{bmatrix}3&4 \\4&-3\end{bmatrix}\begin{bmatrix}x \\y\end{bmatrix}=\begin{bmatrix}24 \\7\end{bmatrix}$$

$$\begin{bmatrix}3x+4y \\4x-3y\end{bmatrix}=\begin{bmatrix}24 \\7\end{bmatrix}$$

$$\therefore 3x+4y=24$$ and $$4x-3y=7$$

Solving the above equations, we get
$$x=4 \ , \ y=3$$

$$\therefore X= \begin{bmatrix}4 \\3\end{bmatrix}$$

Option B.
Question 4
1 / -0

The inverse of $$\begin{bmatrix} 1 & a & b \\ 0 & x & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ is $$\begin{bmatrix} 1 & -a & -b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ then $$x=$$

A
$$a$$

B
$$b$$

C
$$0$$

D
$$1$$

Solution

$$\begin{array}{l} \left[ { \begin{array} { *{ 20 }{ c } }1 & a & b \\ 0 & x & 0 \\ 0 & 0 & 1 \end{array} } \right] \left[ { \begin{array} { *{ 20 }{ c } }1 & { -a } & { -b } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} } \right] =\left[ { \begin{array} { *{ 20 }{ c } }1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} } \right] \\ =\left[ { \begin{array} { *{ 20 }{ c } }1 & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & 1 \end{array} } \right] =\left[ { \begin{array} { *{ 20 }{ c } }1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} } \right] \\ \Rightarrow x=1 \end{array}$$
Question 5
1 / -0

If $$A = \left[ {\begin{array}{*{20}{c}} 2&{ - 3} \\ { - 4}&1 \end{array}} \right]$$, then $$\left[ {3{A^2} + 12A} \right]$$ is equal to

A
$$\left[ {\begin{array}{*{20}{c}}
{72}&{ - 84} \\
{ - 63}&{51}
\end{array}} \right]$$

B
$$\left[ {\begin{array}{*{20}{c}}
{51}&{63} \\
{84}&{72}
\end{array}} \right]$$

C
$$\left[ {\begin{array}{*{20}{c}}
{51}&{84} \\
{63}&{72}
\end{array}} \right]$$

D
$$\left[ {\begin{array}{*{20}{c}}
{72}&{ - 63} \\
{ - 84}&{51}
\end{array}} \right]$$

Solution

We have,
$$A=\left[ \begin{matrix} 2 & -3 \\ -4 & 1 \end{matrix} \right] $$

$$A^2 = A \times A =\left[ \begin{matrix} 2 & -3 \\ -4 & 1 \end{matrix} \right] \times \left[ \begin{matrix} 2 & -3 \\ -4 & 1 \end{matrix} \right] $$
$$A^2=\left[ \begin{matrix} 16 & -9 \\ -12 & 13 \end{matrix} \right] $$

Thus,
$$3A^2=\left[ \begin{matrix} 48 & -27 \\ -36 & 39 \end{matrix} \right] $$

Now, $$12A=\left[ \begin{matrix} 24 & -36 \\ -48 & 12 \end{matrix} \right] $$

Therefore,
$$[3A^2 + 12A]=\left[ \begin{matrix} 48 & -27 \\ -36 & 39 \end{matrix} \right] +\left[ \begin{matrix} 24 & -36 \\ -48 & 12 \end{matrix} \right] \\ [3A^2 +12A]= \left[ \begin{matrix} 72 & -63 \\ -84 & 51 \end{matrix} \right] $$
Question 6
1 / -0

$$A=\begin{bmatrix} a & b \\ 0 & c \end{bmatrix}$$ then $${A}^{-1}+(A-aI)(A-cI)=$$

A
$$\cfrac { 1 }{ ac } \begin{bmatrix} a & b \\ 0 & -c \end{bmatrix}$$

B
$$\cfrac { 1 }{ ac } \begin{bmatrix} -a & b \\ 0 & c \end{bmatrix}$$

C
$$\cfrac { 1 }{ ac } \begin{bmatrix} c & -b \\ 0 & a \end{bmatrix}$$

D
$$\cfrac { 1 }{ ac } \begin{bmatrix} c & b \\ 0 & a \end{bmatrix}$$

Solution

$$A=\begin{bmatrix}a&b\\0&c\end{bmatrix}$$
$$A^{-1}=\dfrac{1}{ac-0(b)}\begin{bmatrix}c&-b\\0&a\end{bmatrix}=\dfrac{1}{ac}\begin{bmatrix}c&-b\\0&a\end{bmatrix}$$
$$(A-aI)(A-cI)=A^{2}-(a+c)A+acI=\begin{bmatrix}a&b\\0&c\end{bmatrix}\begin{bmatrix}a&b\\0&c\end{bmatrix}-(a+c)\begin{bmatrix}a&b\\0&c\end{bmatrix}+acI=\begin{bmatrix}0&0\\0&0\end{bmatrix}$$
$$A^{-1}+(A-{a}I)(A-{c}I)=\dfrac{1}{ac}\begin{bmatrix}c&-b\\0&a\end{bmatrix}$$
Question 7
1 / -0

If $$A=\begin{bmatrix} \cos { x } & \sin { x } \\ -\sin { x } & \cos { x } \end{bmatrix}$$ and $$A(AdjA)=k\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ then the value of $$k$$ is

A
$$\sin{x}\cos{x}$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

$$A(AdjA)=\begin{bmatrix}c^{2}+s^{2}&cs-cs\\-cs+cs&c^{2}+s^{2}\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\implies k=1$$
Question 8
1 / -0

If $$A=\begin{bmatrix} 0 & a+1 & b-2 \\ 2a-1 & 0 & c-2 \\ 2b+1 & 2+c & 0 \end{bmatrix}$$ is skew symmetric then $$a+b+c$$=

A
$$3$$

B
$$-3$$

C
$$\cfrac{1}{3}$$

D
$$-\cfrac{1}{3}$$

Solution

$$A^{T}+A=0\implies a+1+2a-1=0;b-2+2b+1=0;c+2+c-2=0\implies a=c=0,$$
$$b=\displaystyle \frac{1}{3}\implies a+b+c=\frac{1}{3}$$
Question 9
1 / -0

If $$A$$ is square matrix such that $${A^2} = 1$$ then $${A^{ - 1}} = ?$$

A
$$2$$A

B
A

C
$$0$$

D
A+$$1$$

Solution

$${ A }^{ 2 }={ 1 }$$ can be witten as

$$ { A }^{ 2 }={ I }$$
multiply $${ { A }^{ -1 } }$$ on both side
$$\\ { { A }^{ -1 } }\times ({ A }\times { A })={ A }^{ -1 }\times { I }\\ ({ A }^{ -1 }\times { A })\times { A }={ A }^{ -1 }\\ { I }\times { A }={ { A }^{ -1 } }\\ { A }^{ -1 }={ A }\\ $$
Corrrect option is B
Question 10
1 / -0

If $$A=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}$$ then $$A^ {2}$$ is equal to

A
$$A$$

B
$$-A$$

C
$$2A$$

D
$$-2A$$

Solution

$$A^2= \begin{bmatrix} 1 & 0 & 1\\ 0 & 0 & 0\\ 1 & 0 & 1\end{bmatrix} \times \begin{bmatrix} 1 & 0 & 1\\ 0 & 0 & 0\\ 1 & 0 & 1\end{bmatrix}$$

$$A^2=\begin{bmatrix} 2 & 0 & 2\\ 0 & 0 & 0\\ 2 & 0 & 2\end{bmatrix}$$

$$A^2=2\begin{bmatrix} 1 & 0 & 1\\ 0 & 0 & 0\\ 1 & 0 & 1\end{bmatrix}=2A$$

$$\therefore A^2=2A$$.

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Re-Attempt Weekly Quiz Competition

Matrices Test - 23

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Matrices Test - 23

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Question 1

$$0$$

$$3$$

$$6$$

$$8$$

Solution

Question 2

$$1\times 3$$

$$2\times 3$$

$$3\times 3$$

$$2\times 2$$

Solution

Question 3

$$\begin{bmatrix} -4 \\ -3\end{bmatrix}$$

$$\begin{bmatrix} 4 \\ 3\end{bmatrix}$$

$$\begin{bmatrix} -4 \\ 3\end{bmatrix}$$

$$\begin{bmatrix} 4 \\ -3\end{bmatrix}$$

Solution

Question 4

$$a$$

$$b$$

$$0$$

$$1$$

Solution

Question 5

$$\left[ {\begin{array}{*{20}{c}} {72}&{ - 84} \\ { - 63}&{51} \end{array}} \right]$$

$$\left[ {\begin{array}{*{20}{c}} {51}&{63} \\ {84}&{72} \end{array}} \right]$$

$$\left[ {\begin{array}{*{20}{c}} {51}&{84} \\ {63}&{72} \end{array}} \right]$$

$$\left[ {\begin{array}{*{20}{c}} {72}&{ - 63} \\ { - 84}&{51} \end{array}} \right]$$

Solution

Question 6

$$\cfrac { 1 }{ ac } \begin{bmatrix} a & b \\ 0 & -c \end{bmatrix}$$

$$\cfrac { 1 }{ ac } \begin{bmatrix} -a & b \\ 0 & c \end{bmatrix}$$

$$\cfrac { 1 }{ ac } \begin{bmatrix} c & -b \\ 0 & a \end{bmatrix}$$

$$\cfrac { 1 }{ ac } \begin{bmatrix} c & b \\ 0 & a \end{bmatrix}$$

Solution

Question 7

$$\sin{x}\cos{x}$$

$$1$$

$$-1$$

$$2$$

Solution

Question 8

$$3$$

$$-3$$

$$\cfrac{1}{3}$$

$$-\cfrac{1}{3}$$

Solution

Question 9

$$2$$A

A

$$0$$

A+$$1$$

Solution

Question 10

$$A$$

$$-A$$

$$2A$$

$$-2A$$

Solution

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$$\left[ {\begin{array}{*{20}{c}}
{72}&{ - 84} \\
{ - 63}&{51}
\end{array}} \right]$$

$$\left[ {\begin{array}{*{20}{c}}
{51}&{63} \\
{84}&{72}
\end{array}} \right]$$

$$\left[ {\begin{array}{*{20}{c}}
{51}&{84} \\
{63}&{72}
\end{array}} \right]$$

$$\left[ {\begin{array}{*{20}{c}}
{72}&{ - 63} \\
{ - 84}&{51}
\end{array}} \right]$$